1.74 On a training flight, a student pilot flies from Lincoln, Nebraska, to Clarinda, Iowa, then to St. Joseph, Missouri, and then to Manhattan, Kansas (Fig. P1.74).

1.74 On a training flight, a student pilot flies from Lincoln, Nebraska, to Clarinda, Iowa, then to St. Joseph, Missouri, and then to Manhattan, Kansas (Fig. P1.74). The directions are shown relative to north: $0^o$ is north, $90^o$ is east, $180^o$ is south, and $270^o$ is west. Use the method of components to find (a) the distance she has to fly from Manhattan to get back to Lincoln, and (b) the direction (relative to north) she must fly to
get there. Illustrate your solutions with a vector diagram.

SOLUTION

Make a sketch to assess the situation like that of Figure 1.74a. Our +x-axis is pointing north and +y-axis is pointing east. Since the given angles are measured from +x-axis and rotates towards +y-axis then the components are,

$A_x= (147 \ km) \ cos \ 85^o=12.81 \ km$

$A_y= (147 \ km) \ sin \ 85^o=146.4 \ km$

$B_x= (106 \ km) \ cos \ 167^o=-103.3 \ km$

$B_y= (106 \ km) \ sin \ 167^o=23.84 \ km$

$C_x= (166 \ km) \ cos \ 235^o=-95.21 \ km$

$C_y= (166 \ km) \ sin \ 235^o=-136.0 \ km$

Since the student pilot will go back to where he started, then let his last displacement be vector $\vec{D}=D_x+D_y$ with components and also his total displacement $\vec{R}$ is zero,

$\vec{R}=\vec{A}+\vec{B}+\vec{C}+\vec{D}=0$ then rearranging it we have, $\vec{D}=-\vec{A}-\vec{B}-\vec{C}$. Its components are,

$D_x=-A_x-B_x-C_x=-12.81 \ km-(-103.3 \ km)-(-95.21 \ km)$

$=185.7 \ km$

$D_y=-A_y-B_y-C_y=-146.4 \ km-23.84 \ km-(-136.0 \ km)$

$=-34.24 \ km$

then its magnitude and direction are

$D=\sqrt{(D_x)^2+(D_y)^2}=189 \ km$

$tan \ \theta=\frac{D_y}{D_x}$

$\theta=-10^o$ measured counterclockwise from north direction, or $10^o$ west of north.

The vector diagram below shows a rough sketch of $\vec{A}+\vec{B}+\vec{C}+\vec{D}=0$